leetcode 105 construct binary tree from preorder and inorder traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

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preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

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  3
/
9 20
/
15 7

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import utils.TreeNode;

import java.util.Arrays;

public class {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return helper(preorder, inorder, 0, preorder.length, 0, inorder.length);
}
public TreeNode helper(int[] preorder, int[] inorder, int prel, int prer, int inl ,int inr) {
if (inl == inr) return new TreeNode(inorder[inl]);
TreeNode root = new TreeNode(preorder[prel]);
int i = inl;
for (i = inl; i<inr; i++) {
if (inorder[i] == root.val) {
break;
}
}
root.left = helper(preorder, inorder, prel+1,prel+(i-inl), inl, i-1);
root.right = helper(preorder, inorder, prel+(i-inl)+1, prer, i+1, inr);
return root;
}
}